Question

If \(\int \frac{2^{1/x}}{x^2} \, dx = k\,2^{1/x} + c\) then \(k\) is equal to

• A. $\frac{1}{\log 2}$
• B. $\log 4$
• C. $\frac{1}{\log 3}$
• D. $-\frac{1}{\log 3}$
• E. $-\frac{1}{\log 2}$

Hint:

For expressions like $a^{1/x}$: - Use chain rule carefully - $\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$ - Match the derivative form to identify constants quickly

Answer 1

Answer: (E)

Concept: We use reverse differentiation. If: \[ \int g(x)\,dx = F(x) + c \] then: \[ g(x) = \frac{d}{dx}[F(x)] \] We differentiate $2^{1/x}$ using chain rule: \[ \frac{d}{dx}\left(a^u\right) = a^u \ln a \cdot \frac{du}{dx} \]

Step 1: Differentiate $2^{1/x}$.
\[ \frac{d}{dx}\left(2^{1/x}\right) = 2^{1/x} \ln 2 \cdot \frac{d}{dx}\left(\frac{1}{x}\right) \] \[ = 2^{1/x} \ln 2 \cdot \left(-\frac{1}{x^2}\right) \] \[ = -\frac{2^{1/x} \ln 2}{x^2} \]

Step 2: Compare with given integrand.
\[ \frac{2^{1/x}}{x^2} \] We observe: \[ \frac{2^{1/x}}{x^2} = -\frac{1}{\ln 2} \cdot \frac{d}{dx}\left(2^{1/x}\right) \]

Step 3: Integrate both sides.
\[ \int \frac{2^{1/x}}{x^2} \, dx = -\frac{1}{\ln 2} \cdot 2^{1/x} + c \]

Step 4: Compare with given form $k\,2^{1/x} + c$.
\[ k = -\frac{1}{\ln 2} \]